Problem: $\begin{aligned} y&=(2x^2-1)^6 \\\\ \dfrac{dy}{dx}&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $24x^{11}$ (Choice B) B $6(4x)^5$ (Choice C) C $24x(2x^2-1)^5$ (Choice D, Checked) D $6(2x^2-1)^5$
Explanation: Since $(2x^2-1)^6$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $\underbrace{(~\overbrace{2x^2-1}^{\text{inner}}~)^6}_{\text{outer}}$ So if $(2x^2-1)^6=w(u(x))$, then: $\begin{aligned} {u(x)}&={2x^2-1} &&\text{inner function} \\\\ w(x)&=x^6&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={4x} \\\\ {w'(x)}&={6x^5} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={6({2x^2-1})^5} \cdot {4x} \\\\ &=24x(2x^2-1)^5 \end{aligned}$